Singular Value Decomposition (SVD). Show equality.

I have to show the following equality:

$WW^{T} = USUS$, where $W = USV^{T}$; $U,V$ are orthogonal matrices and $S$ is a diagonal matrix, where all entries but the diagonal are $0$.

I think I am missing just one step to finish the proof. I am this far: $WW^{T} = (USV^T)(USV^T)^T = (USV^T)(VS^TU^T) = USV^TVS^TU^T = USS^TU^T$. Then, since $S = S^T$, it follows that $USS^TU^T = USSU^T$.

Now my questions is, is the equation $SU^T = US$ valid, and if so, why? As I am not a mathematician, my knowledge when it comes to Linear Algebra is not the best. So sorry, if this is a rather trivial question.

Thanks in advance.

1 Answer

It is not true.

Let $U=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, $S = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$, $V = I$, which gives $W=\begin{bmatrix} 0 & 2 \\ 1 & 0 \end{bmatrix}$.

A quick computation shows $W W^T = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix}$, $USUS = 2 I$.

• Thanks for your quick response. Hmm...then there must be something wrong with the exercise specification. Does this mean that $USSU^T$ is already the result? – trojan Jan 17 '14 at 18:32
• Well, it is true from direct computation that $U S^2 U^T = W W^T$. – copper.hat Jan 17 '14 at 18:41
• of course, you are right. – trojan Jan 17 '14 at 19:06